Question: $\dfrac{d}{dx}(-2x^5-3x^3+1)=$
Explanation: According to the sum rule, the derivative of $-2x^5-3x^3+1$ is the sum of the derivatives of $-2x^5$, $-3x^3$, and $1$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(-2x^5)&=-2\dfrac{d}{dx}(x^5)&&\gray{\text{Constant multiple rule}}\\\\ &=-2\cdot (5x^4)&&\gray{\text{Power rule}}\\ \\ &=-10x^4\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-2x^5-3x^3+1) \\\\ &=-2\dfrac{d}{dx}(x^5)-3\dfrac{d}{dx}(x^3)+\dfrac{d}{dx}(1)&&\gray{\text{Basic differentiation rules}} \\\\ &=-2\dfrac{d}{dx}(x^5)-3\dfrac{d}{dx}(x^3)+0&&\gray{\text{Constant rule}} \\\\ &=-2\cdot 5x^4-3\cdot3x^2&&\gray{\text{The power rule}} \\\\ &=-10x^4-9x^2 \end{aligned}$ In conclusion, $\dfrac{d}{dx}(-2x^5-3x^3+1)=-10x^4-9x^2$.